CodeForces - 220B Little Elephant and Array （莫队+离散化）-优快云博客

本文解决了一个编程挑战，小象与数组的问题，涉及数组操作和查询处理。通过离散化处理和Mo's算法，有效地解决了在给定区间内寻找出现次数等于其值的元素数量的问题。

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Little Elephant and Array

The Little Elephant loves playing with arrays. He has array a, consisting of npositive integers, indexed from 1 to n. Let's denote the number with index i as ai.

Additionally the Little Elephant has m queries to the array, each query is characterised by a pair of integers lj and rj (1 ≤ lj ≤ rj ≤ n). For each query lj, rjthe Little Elephant has to count, how many numbers x exist, such that number xoccurs exactly x times among numbers alj, alj + 1, ..., arj.

Help the Little Elephant to count the answers to all queries.

Input

The first line contains two space-separated integers n and m (1 ≤ n, m ≤ 105) — the size of array a and the number of queries to it. The next line contains n space-separated positive integers a1, a2, ..., an (1 ≤ ai ≤ 109). Next m lines contain descriptions of queries, one per line. The j-th of these lines contains the description of the j-th query as two space-separated integers lj and rj (1 ≤ lj ≤ rj ≤ n).

Output

In m lines print m integers — the answers to the queries. The j-th line should contain the answer to the j-th query.

Examples

Input

7 2
3 1 2 2 3 3 7
1 7
3 4

Output

3
1

题目链接：http://codeforces.com/problemset/problem/220/B

题目大意：一个长度为n的数列，m次查询，每次查询求出区间 l 到 r 区间内有几个数 x 在区间内出现了x次

思路：因为ai最大到了10^9没办法标记，所以先对数组a进行离散化处理，然后就是莫队模板题

代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<math.h>
using namespace std;
const int N=200010;
int a[N],ans,c[N],b[N],e[N],k;
struct node
{
    int l,r,id,pos;
}q[N];
bool cmp(node a,node b)
{
    if(a.pos==b.pos) return a.r<b.r;
    return a.pos<b.pos;
}
void update(int x,int p)
{
    if(x==0) return ;
    int y=lower_bound(e+1,e+1+k,a[x])-e;
    if(c[y]==a[x]) ans-=1;
    c[y]+=p;
    if(c[y]==a[x]) ans+=1;
}
int main()
{
        int n,m;
        scanf("%d%d",&n,&m);
        int p=sqrt(n);
        for(int i=1;i<=n;i++) scanf("%d",&a[i]),e[i]=a[i];
        sort(e+1,e+1+n);
        k=unique(e+1,e+n+1)-(e+1);
        for(int i=0;i<m;i++)
        {
            scanf("%d%d",&q[i].l,&q[i].r);
            q[i].id=i;
            q[i].pos=q[i].l/p;
        }
        sort(q,q+m,cmp);
        int l=0,r=0;
        ans=0;
        for(int i=0;i<m;i++)
        {
            while(l<q[i].l) update(l++,-1);
            while(l>q[i].l) update(--l,1);
            while(r>q[i].r) update(r--,-1);
            while(r<q[i].r) update(++r,1);
            b[q[i].id]=ans;
        }
        for(int i=0;i<m;i++)
            printf("%d\n",b[i]);
    return 0;
}